How long does it take for the car to pass the truck?

AI Thread Summary
The discussion revolves around a physics problem where a car attempts to pass a truck while considering acceleration and the distance of an oncoming vehicle. The car accelerates at 1.0 m/s² and needs to cover a total distance of 40 meters to safely pass the truck. Calculations indicate that it takes 8.9 seconds and 262.1 meters to complete the maneuver, but there are discrepancies when comparing distances with the approaching car's position. Suggestions for improvement include clearly labeling each step of the calculations and using position equations for all vehicles involved to determine the time of passing more intuitively. The conversation emphasizes the importance of correctly applying kinematic equations to solve the problem.
physicsguy69
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Homework Statement



a car is behind a truck going 25m/s on the highway. the car's river looks for a chance to pass, guessing that his car can accelerate at 1.0m/s^2. he gauges that he has to cover the 20 meters length of the truck 10 meters clear room at the rear of the truck and 10 meters more at the front of it. in the oncoming lane, he sees a car approaching, probably also traveling at 25m/s. he estimates that the car is about 400m away

Homework Equations



v^2=Vo^2+2ax
v=Vo+at
d=rt

The Attempt at a Solution



v^2=(2)(1)(40) V=8.9

8.9+25=33.9m/s
1149.2=625+2(1)(x) x=262.1m
it takes car one 262.1meters to pass the truck and this process takes 8.9 seconds

d=rt
d=(8.9sec)(25m/s) d=222.5m

400 - 262.1 ≠ 222.5
400 - 262.1 =137.9

final answer = no

is there anything wrong here?
 
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physicsguy69 said:

Homework Statement



a car is behind a truck going 25m/s on the highway. the car's river looks for a chance to pass, guessing that his car can accelerate at 1.0m/s^2. he gauges that he has to cover the 20 meters length of the truck 10 meters clear room at the rear of the truck and 10 meters more at the front of it. in the oncoming lane, he sees a car approaching, probably also traveling at 25m/s. he estimates that the car is about 400m away

Homework Equations



v^2=Vo^2+2ax
v=Vo+at
d=rt

The Attempt at a Solution



v^2=(2)(1)(40) V=8.9

8.9+25=33.9m/s
1149.2=625+2(1)(x) x=262.1m
it takes car one 262.1meters to pass the truck and this process takes 8.9 seconds

d=rt
d=(8.9sec)(25m/s) d=222.5m

400 - 262.1 ≠ 222.5
400 - 262.1 =137.9

final answer = no

is there anything wrong here?

Welcome to the PF.

It looks like you are taking a reasonable approach, but it is a little hard to follow your reasoning. If you could label what you are doing in each step, that would help.

Also, I would (personally) approach it a little differently. I would write 3 equations for the positions of each of the 3 vehicles, something like

Xc1(t) =
Xc2(t) =
Xt(t) =

Use the initial conditions you are given in the problem (like initial positions and speeds), and then I would solve for the time t when the two cars pass each other. And given that time t, would the first car be the 10m past the truck or not. Doing it that way is more intuitive to me, but different folks will approach problems in different ways.

Still, you could use my approach to check your answer...
 
what would your equations be?
 
physicsguy69 said:
what would your equations be?

What is the basic kinematic equation for situations where there is a constant accelertaion (like questions involving gravitational acceleration, or this problem with the constant car acceleration)?

The basic equation is for the distance as a function of time x(t), in terms of the initial position, initial velocity, acceleration, and time. Can you write that general equation? And then use it 3 times, using the appropriate Xo, Vo, etc. for each of the 3 vehicles...
 
x=1/2at^2 but the truck and the other car have a constant velocity
 
physicsguy69 said:
x=1/2at^2 but the truck and the other car have a constant velocity

That's not the whole equation. You left off the Xo and Vo terms...
 
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